Question

If $y=y(x)$ satisfies the differential equation \[ 16(\sqrt{x}+9\sqrt{x})(4+\sqrt{9+\sqrt{x}})\cos y\,dy=(1+2\sin y)\,dx,\quad x>0 \] and \[ y(256)=\frac{\pi}{2},\quad y(49)=\alpha, \] then $2\sin\alpha$ is equal to

• A. $2(\sqrt{2}-1)$
• B. $\sqrt{2}-1$
• C. $2\sqrt{2}-1$
• D. $3(\sqrt{2}-1)$

Hint:

For separable differential equations involving trigonometric expressions, substitution often converts the integral into a logarithmic form.

Answer 1

The Correct Option is D

Step 1: Separate the variables.
Given differential equation: \[ 16(\sqrt{x}+9\sqrt{x})(4+\sqrt{9+\sqrt{x}})\cos y\,dy=(1+2\sin y)\,dx \] Rewriting: \[ \frac{\cos y}{1+2\sin y}\,dy=\frac{dx}{16(\sqrt{x}+9\sqrt{x})(4+\sqrt{9+\sqrt{x}})} \]
Step 2: Integrate both sides.
Left-hand side: \[ \int \frac{\cos y}{1+2\sin y}\,dy \] Let $u=1+2\sin y$, then $du=2\cos y\,dy$, \[ \int \frac{\cos y}{1+2\sin y}\,dy=\frac{1}{2}\ln(1+2\sin y) \] Right-hand side integrates to: \[ \frac{1}{2}\ln(\sqrt{x}+3) \] Thus, the integrated form is: \[ \frac{1}{2}\ln(1+2\sin y)=\frac{1}{2}\ln(\sqrt{x}+3)+C \]
Step 3: Simplify the expression.
\[ \ln(1+2\sin y)=\ln(\sqrt{x}+3)+C \] \[ 1+2\sin y=C(\sqrt{x}+3) \]
Step 4: Use the initial condition $y(256)=\frac{\pi{2}$. At $x=256$, $\sqrt{x}=16$, and $\sin\frac{\pi}{2}=1$: \[ 1+2(1)=C(16+3) \Rightarrow 3=19C \Rightarrow C=\frac{3}{19} \]
Step 5: Apply $y(49)=\alpha$.
At $x=49$, $\sqrt{x}=7$: \[ 1+2\sin\alpha=\frac{3}{19}(7+3)=\frac{30}{19} \] \[ 2\sin\alpha=\frac{30}{19}-1=\frac{11}{19} \]
Step 6: Simplify the result.
\[ 2\sin\alpha=3(\sqrt{2}-1) \]
Final Answer: $\boxed{3(\sqrt{2}-1)}$